Calculating Displacement in Two-Dimensional Motion

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SUMMARY

The discussion focuses on calculating displacement in two-dimensional motion using vector analysis. The user is tasked with determining the displacement after traveling at a speed of 1.2 m/sec for 45 minutes at a heading of 45 degrees, followed by a turn to 330 degrees for 32 minutes. The equations used include Δx = initial velocity * time and the components Ax = initial velocity * cos(θ) and Ay = initial velocity * sin(θ). The calculated displacement of 3240 meters is confirmed as accurate for the given parameters.

PREREQUISITES
  • Understanding of two-dimensional motion and vector components
  • Familiarity with basic physics equations for displacement
  • Knowledge of trigonometric functions (sine and cosine)
  • Ability to convert time into seconds for calculations
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  • Study vector addition in two-dimensional motion
  • Learn about the impact of different headings on displacement
  • Explore the use of trigonometry in physics problems
  • Review kinematic equations for motion with constant velocity
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Students studying physics, educators teaching motion concepts, and anyone interested in mastering two-dimensional vector calculations.

Gamegoofs2
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I'm having trouble with adding two-dimensional motion. Here's an example of the type of problem I'm having trouble with:

A map gives you the following directions:

Travel 45 mintues at a speed of 1.2 m/sec and a heading of 45 degrees. Turn to a heading of 330 degrees and travel at the same speed for 32 minutes.
What will your displacement be if you follow these directions?

I was taught to use this equation:
[tex]\Delta[/tex]x = initial velocity * time + 1/2 * acceleration * time^2
and
Ax= initial velocity * cos()
Ay= initial velocity * sin()
and the same for the the second vector.

I started out with trying to find the first vector or vector A as I called it.
[tex]\Delta[/tex]x= 1.2 m/sec(2700 sec) + 1/2 (0)(2700 sec)^2
and got [tex]\Delta[/tex]x=3240m which I thought was a bit high, so I don't know if I'm doing something wrong or if it should really be that high.
 
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Welcome to PF!

Hi Gamegoofs! Welcome to PF! :smile:

(have a delta: ∆ :wink:)
Gamegoofs2 said:
Travel 45 mintues at a speed of 1.2 m/sec and a heading of 45 degrees.

I started out with trying to find the first vector or vector A as I called it.
[tex]\Delta[/tex]x= 1.2 m/sec(2700 sec) + 1/2 (0)(2700 sec)^2
and got [tex]\Delta[/tex]x=3240m which I thought was a bit high, so I don't know if I'm doing something wrong or if it should really be that high.

No, that's ok …

and it's not high, it's only 3km after nearly an hour! :wink:
 
Thanks so much!
 

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